Integrand size = 14, antiderivative size = 98 \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=-\frac {b^2 \cot (c+d x) \log (\cos (c+d x)) \sqrt {b \tan ^2(c+d x)}}{d}-\frac {b^2 \tan (c+d x) \sqrt {b \tan ^2(c+d x)}}{2 d}+\frac {b^2 \tan ^3(c+d x) \sqrt {b \tan ^2(c+d x)}}{4 d} \]
-b^2*cot(d*x+c)*ln(cos(d*x+c))*(b*tan(d*x+c)^2)^(1/2)/d-1/2*b^2*(b*tan(d*x +c)^2)^(1/2)*tan(d*x+c)/d+1/4*b^2*(b*tan(d*x+c)^2)^(1/2)*tan(d*x+c)^3/d
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.57 \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=-\frac {\cot (c+d x) \left (-1+2 \cot ^2(c+d x)+4 \cot ^4(c+d x) \log (\cos (c+d x))\right ) \left (b \tan ^2(c+d x)\right )^{5/2}}{4 d} \]
-1/4*(Cot[c + d*x]*(-1 + 2*Cot[c + d*x]^2 + 4*Cot[c + d*x]^4*Log[Cos[c + d *x]])*(b*Tan[c + d*x]^2)^(5/2))/d
Time = 0.40 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.68, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4141, 3042, 3954, 3042, 3954, 3042, 3956}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \tan (c+d x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \int \tan ^5(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \int \tan (c+d x)^5dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \left (\frac {\tan ^4(c+d x)}{4 d}-\int \tan ^3(c+d x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \left (\frac {\tan ^4(c+d x)}{4 d}-\int \tan (c+d x)^3dx\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \left (\int \tan (c+d x)dx+\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \left (\int \tan (c+d x)dx+\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle b^2 \cot (c+d x) \sqrt {b \tan ^2(c+d x)} \left (\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}-\frac {\log (\cos (c+d x))}{d}\right )\) |
b^2*Cot[c + d*x]*Sqrt[b*Tan[c + d*x]^2]*(-(Log[Cos[c + d*x]]/d) - Tan[c + d*x]^2/(2*d) + Tan[c + d*x]^4/(4*d))
3.1.24.3.1 Defintions of rubi rules used
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Ta n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Time = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59
method | result | size |
derivativedivides | \(\frac {{\left (b \left (\tan ^{2}\left (d x +c \right )\right )\right )}^{\frac {5}{2}} \left (\tan ^{4}\left (d x +c \right )-2 \left (\tan ^{2}\left (d x +c \right )\right )+2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )\right )}{4 d \tan \left (d x +c \right )^{5}}\) | \(58\) |
default | \(\frac {{\left (b \left (\tan ^{2}\left (d x +c \right )\right )\right )}^{\frac {5}{2}} \left (\tan ^{4}\left (d x +c \right )-2 \left (\tan ^{2}\left (d x +c \right )\right )+2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )\right )}{4 d \tan \left (d x +c \right )^{5}}\) | \(58\) |
risch | \(\frac {b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, x}{{\mathrm e}^{2 i \left (d x +c \right )}-1}-\frac {2 b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left (d x +c \right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) d}-\frac {4 i b^{2} \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{6 i \left (d x +c \right )}+{\mathrm e}^{4 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} d}-\frac {i b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) d}\) | \(300\) |
1/4/d*(b*tan(d*x+c)^2)^(5/2)*(tan(d*x+c)^4-2*tan(d*x+c)^2+2*ln(1+tan(d*x+c )^2))/tan(d*x+c)^5
Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {{\left (b^{2} \tan \left (d x + c\right )^{4} - 2 \, b^{2} \tan \left (d x + c\right )^{2} - 2 \, b^{2} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 3 \, b^{2}\right )} \sqrt {b \tan \left (d x + c\right )^{2}}}{4 \, d \tan \left (d x + c\right )} \]
1/4*(b^2*tan(d*x + c)^4 - 2*b^2*tan(d*x + c)^2 - 2*b^2*log(1/(tan(d*x + c) ^2 + 1)) - 3*b^2)*sqrt(b*tan(d*x + c)^2)/(d*tan(d*x + c))
\[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\int \left (b \tan ^{2}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.48 \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\frac {b^{\frac {5}{2}} \tan \left (d x + c\right )^{4} - 2 \, b^{\frac {5}{2}} \tan \left (d x + c\right )^{2} + 2 \, b^{\frac {5}{2}} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{4 \, d} \]
1/4*(b^(5/2)*tan(d*x + c)^4 - 2*b^(5/2)*tan(d*x + c)^2 + 2*b^(5/2)*log(tan (d*x + c)^2 + 1))/d
Leaf count of result is larger than twice the leaf count of optimal. 646 vs. \(2 (88) = 176\).
Time = 1.18 (sec) , antiderivative size = 646, normalized size of antiderivative = 6.59 \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\text {Too large to display} \]
-1/4*(2*b^2*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^ 2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*sgn(tan(d*x + c))*tan(d*x)^4*tan( c)^4 + 3*b^2*sgn(tan(d*x + c))*tan(d*x)^4*tan(c)^4 - 8*b^2*log(4*(tan(d*x) ^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + t an(c)^2 + 1))*sgn(tan(d*x + c))*tan(d*x)^3*tan(c)^3 + 2*b^2*sgn(tan(d*x + c))*tan(d*x)^4*tan(c)^2 - 8*b^2*sgn(tan(d*x + c))*tan(d*x)^3*tan(c)^3 + 2* b^2*sgn(tan(d*x + c))*tan(d*x)^2*tan(c)^4 + 12*b^2*log(4*(tan(d*x)^2*tan(c )^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*sgn(tan(d*x + c))*tan(d*x)^2*tan(c)^2 - b^2*sgn(tan(d*x + c))*tan(d* x)^4 - 8*b^2*sgn(tan(d*x + c))*tan(d*x)^3*tan(c) + 4*b^2*sgn(tan(d*x + c)) *tan(d*x)^2*tan(c)^2 - 8*b^2*sgn(tan(d*x + c))*tan(d*x)*tan(c)^3 - b^2*sgn (tan(d*x + c))*tan(c)^4 - 8*b^2*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*ta n(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*sgn(tan(d*x + c))*tan(d*x)*tan(c) + 2*b^2*sgn(tan(d*x + c))*tan(d*x)^2 - 8*b^2*sgn(tan( d*x + c))*tan(d*x)*tan(c) + 2*b^2*sgn(tan(d*x + c))*tan(c)^2 + 2*b^2*log(4 *(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan( d*x)^2 + tan(c)^2 + 1))*sgn(tan(d*x + c)) + 3*b^2*sgn(tan(d*x + c)))*sqrt( b)/(d*tan(d*x)^4*tan(c)^4 - 4*d*tan(d*x)^3*tan(c)^3 + 6*d*tan(d*x)^2*tan(c )^2 - 4*d*tan(d*x)*tan(c) + d)
Timed out. \[ \int \left (b \tan ^2(c+d x)\right )^{5/2} \, dx=\int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}^{5/2} \,d x \]